Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<int> D[100001], C[100001];
vector<int> M[20][100001], Ve;
int P[20][100001], De[100001], c[100001];
void dfs(int d, int p) {
  for (int i = 0; i < D[d].size(); i++) {
    int a = D[d][i];
    if (a != p) {
      int ca = C[a].size(), cd = C[d].size();
      for (int y = 0; y < min(10, ca); y++) {
        M[0][a].push_back(C[a][y]);
      }
      for (int y = 0; y < min(10, cd); y++) {
        M[0][a].push_back(C[d][y]);
      }
      sort(M[0][a].begin(), M[0][a].end());
      while (M[0][a].size() > 10) {
        M[0][a].pop_back();
      }
      De[a] = De[d] + 1;
      P[0][a] = d;
      dfs(a, d);
    }
  }
}
void merge(int Y, int I) {
  int s = M[Y][I].size();
  for (int i = 0; i < min(10, s); i++) {
    int g = 1;
    for (int j = 0; j < Ve.size(); j++) {
      if (Ve[j] == M[Y][I][i]) {
        g = 0;
        break;
      }
    }
    if (g == 1) {
      Ve.push_back(M[Y][I][i]);
    }
  }
  sort(Ve.begin(), Ve.end());
  while (Ve.size() > 10) {
    Ve.pop_back();
  }
}
int main() {
  int n, m, q, a, b, V, U, A;
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    D[a].push_back(b);
    D[b].push_back(a);
  }
  for (int i = 0; i < m; i++) {
    cin >> c[i];
    C[c[i]].push_back(i + 1);
  }
  for (int i = 0; i < 100001; i++) {
    sort(C[i].begin(), C[i].end());
  }
  De[1] = 0;
  P[0][1] = -1;
  dfs(1, -1);
  for (int i = 1; i < 20; i++) {
    for (int y = 1; y < 100001; y++) {
      if (P[i - 1][y] != -1 && P[i - 1][P[i - 1][y]] != -1) {
        int p = P[i - 1][y];
        P[i][y] = P[i - 1][p];
        int s1 = M[i - 1][y].size(), s2 = M[i - 1][p].size();
        for (int u = 0; u < min(10, s1); u++) {
          M[i][y].push_back(M[i - 1][y][u]);
        }
        sort(M[i][y].begin(), M[i][y].end());
        while (M[i][y].size() > 10) {
          M[i][y].pop_back();
        }
        for (int u = 0; u < min(10, s2); u++) {
          int g = 1;
          for (int j = 0; j < M[i][y].size(); j++) {
            if (M[i][y][j] == M[i - 1][p][u]) {
              g = 0;
              break;
            }
          }
          if (g == 1) {
            M[i][y].push_back(M[i - 1][p][u]);
          }
        }
        sort(M[i][y].begin(), M[i][y].end());
        while (M[i][y].size() > 10) {
          M[i][y].pop_back();
        }
      } else {
        P[i][y] = -1;
      }
    }
  }
  for (int i = 0; i < q; i++) {
    cin >> V >> U >> A;
    if (De[V] > De[U]) {
      swap(V, U);
    }
    if (V == U) {
      int s = C[V].size();
      for (int y = 0; y < min(A, s); y++) {
        Ve.push_back(C[V][y]);
      }
    }
    for (int y = 19; y >= 0; y--) {
      if (P[y][U] != -1 && De[P[y][U]] >= De[V]) {
        merge(y, U);
        U = P[y][U];
      }
    }
    for (int y = 19; y >= 0; y--) {
      if (P[y][U] != -1 && P[y][V] != -1 && P[y][U] != P[y][V]) {
        merge(y, U);
        merge(y, V);
        U = P[y][U];
        V = P[y][V];
      }
    }
    if (U != V) {
      merge(0, U);
      merge(0, V);
    }
    int s = Ve.size();
    cout << min(s, A) << " ";
    for (int y = 0; y < min(s, A); y++) {
      cout << Ve[y] << " ";
    }
    cout << "\n";
    Ve.clear();
  }
  return 0;
}